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User blog:Friendlysociopath/Friendly Does Calcs?
Indeed he does. However, I mainly do them only for myself and don't follow any specific formats or rules, so they may or may not be applicable to other sites; they're just for me and my Deathbattles. However, it's been requested that I post at least a few of them, so I'll leave them here every now and then when I find the time to organize them- in case anyone wants to take a look. I will again state that, these are not made to match the criteria of other sites. As such, it's possible they wouldn't be accepted at some site or another. I do not care. Note: Many of these are parts of conversations and I just copy+pasted them. Expect odd text or formatting unless I'm unusually diligent in making sure they look right. Maka cuts through Gopher's blast (Soul Eater) Probably the most spite-filled calc I've ever done- I was told Maka was only building and decided, 'The hell with you' and whipped up this scene to work with. First, we need our premise, obvious as it might be. http://www.mangareader.net/soul-eater/64/18 Here we see Gopher fire his projectile and destroy two mountain-tops. Interestingly- they appear to be blasted into more or less dust as for the most part there is a giant hole with relatively few stones of decent size where the rock had been. As Maka later cuts directly through a larger blast later- this could be theorized to be the amount of power she can output. In fact, she cuts through it so hard that her attack still manages to hit Gopher and harm him through his armored wings, so well beneath her total power. Next up- how much rock was destroyed? Hard to say as we only see the rock destroyed in the one scan and never again or with more information- but an idea has struck me for a possible method. Here http://www.mangareader.net/soul-eater/64/16 we see Maka is perhaps 3/4ths the size of the number 4. Various scenes in those chapters show trees approaching half the size of the numbers. So if the number was 4 feet tall, Maka would be 3 and the trees would be 2. Maka's stated height is 5'3, 5'4 after the time-skip. The Gopher battle is post time-skip. So Maka's height is 1.626 meters. Divided by .75 = 2.168 meters for the numbers. Half of that is 1.084 meters for the trees. Now, we have to state our assumptions as the other things are relatively easy to visually gauge. The main assumption being to gauge the rock shattered as if it was a perfect cone, the others being that the second mountain-top was the same size and that it's all shattered, not just half and the rest of it falling over. Given the amount of rock we see falling in the second scene, the amount destroyed does not seem remotely unfair as we should be seeing a hella lot more rock falling, and the first scene indicates rather strongly most of the rock is just obliterated. I'd rather not pixel-scale the back mountain like I did the front because it's some distance away and because we don't have a good tree straight-up and visible to start with. Now the scene does not help all that much with gauging the size of the cone. I will admit that and probably be very annoyed by it. However I gave it a short at estimating the site of the mountain tops. As such- 'de image' For reference, the red line is 71 pixels tall, the green line goes from 206 to 884, so 678 pixels along the x axis, and the orange line goes along the y axis from 9 to 1341, so 1332 pixels. The red line is the height of the tree, the green line is the diameter of the bottom of the cone, and the orange line is the height. So 71 pixels = 1.084 meters. Now what we want is the volume of the cone/mountain, which first requires us to know the radius of the circle. 678 / 71 = 9.55x larger than a tree. We then multiply this by our 1.084 to get the diameter, 10.35 meters. Half of that gives us our radius, 5.175 meters. Next we need the height, 1332 / 71 = 18.76x larger than a tree. We then multiply this by our 1.084 to get the height, 20.34 meters. Volume here becomes child's play so long as the cone assumption is accepted, solving for volume gives us 570.43 meters cubed of rock. And so long as the assumption of both mountain-tops being equal is held, that doubles to 1140.86 meters cubed of rock. The Vsbattles site uses values cited in Narutoforums, which are as follows Rock fragmentation=8 j/cc Rock violent fragmentation=69 j/cc Rock Pulverization>200j/cc-214 j/cc Pulverization is smashing the entire thing to dust so it doesn't quite fit that. Violent fragmentation means you can still see pieces but otherwise the debris is too small to make out. Seems perfect. 570.43 meters cubed becomes 570,430,000 cubic centimeters. 69 x 570430000 = 39,359,670,000 Joules or 9.41 tons of tnt. This could be doubled to 18.82 tons of tnt. I know they had the possibility of 120 Joules/cc instead of 69 but choose 69 for lowball purposes. However I happen to think a substantial portion of the mountain being reduced to dust is fair judging by the scans, so I'm going to use 120 for the hell of it. 120 x 570430000 = 68,451,600,000 Joules or 16.36 tons of tnt. Doubled that would be 32.72 tons of tnt. Cloud and Sephiroth Fall (FF7) A video editing exercise also coincidentally used to make a calc. FYI this was hinged around learning how to advance footage in speed so it could only ever be done in multipliers for the program I had at the time, 2x, 4x, 8x, so while the actual final result should only be a 7x change- it is 8x instead. Video can be found here https://www.youtube.com/watch?v=K1wTuAYiVs8 Shinra HQ has 70 floors. Sephiroth causes a portion of the top to fall, mainly the roof since the final battle takes place in the President's office. During the course of Cloud and Sephiroth fighting, it takes over 50 seconds for the various pieces of building to land. The building should've been entirely down in about 7 seconds or so due to gravity. Cloud and Sephiroth are already assumed to be fighting at superhuman speeds as various FF7 characters have blocked and deflected bullets with no issue; meaning we already know they're moving faster than what we're watching. So, since it took 50 seconds for a 7 second event to happen; that would mean what we're watching is only happening around 1/7th the actual speed it should happen in. The video is x8 at the end but it also has 2x and 4x for comparison. 8x is closer to the actual speeds than 4x. Now, what that means is when they cross 1 meter in 1 second of animation, they're actually crossing the 1 meter in 0.143 seconds, so 1 m/s would become 7m/s. Striking 7 times in a second becomes instead becomes 7 times in 0.143 seconds, so 49 times a second. You could in theory try applying this to other battles in the movie but you probably shouldn't as that's not a good assumption to make. Likewise you could argue that Cloud's Omnislash takes place "in the blink of an eye", which, contrary to popular belief, does yield reasonably definite results in a certain area from 100 to 300 milliseconds. As that's 15 strikes in anywhere from 1/10th to 1/3rd of a second, you can easily surpass the 49 times a second mark anyways since it will yield results of 150 and 50 strikes a second respectively. With the use of limit breaks, granted. Zack Spins Around Bullets (FF7) Facing revision! During the events of Last Order, Zack defends Cloud with the Buster Sword while using himself as bait and evading bullets. During this part of the OVA, Zack can be seen moving faster than the bullets. But how fast was he? The rifles are dead-on looking like M16s in Last Order. 960m/s, 2147.46 mph, or 2.8x the speed of sound for bullet velocity. Zack during that scene was moving fast enough that he could spin around practically 180 degrees before the bullets moved a meter. So our timeframe is 1 / 960 seconds, or 0.001 seconds. Average human shoulder width is 17-18 inches, but that's not for athletes and Zack is fairly muscular so we'll go with 50 centimeters. With a radius of 25 centimeters, that gives us a circumference of 157.08 centimeters, 1.5708 meters. Zack traveled half of that, so 0.7854 meters in 0.001 seconds. This gives us a value of Mach 2.2 However, that used a value intended to give a lowball. Namely, saying the bullets couldn't travel a full meter. We can see Zack is exceeding the bullets in speed so this could also be argued to be actually wrong. It's not farfetched to insist the bullets could not travel a foot, which, would multiply the result by 3.3x to give us higher than Mach 7. To be moving his torso at those speeds- he would have to move his legs and arms at comparable or even higher speeds. Midgar Zolom Size Since it's a relatively early feat I thought it should be calc'd since it's not that hard to get a rough idea of how heavy the thing is. You know the score, after narrowly avoiding the seemingly unbeatable Midgar Zolom, Cloud and company get to the edge of the swamp to find... Sephiroth had no such issue dealing with it. A snake is found seemingly impaled atop a tree- so how heavy is that beastie? (Eyeball estimate) AFAIK snakes only a foot thick and in the teens for length can weight 500+ pounds. If you multiply this by 8 for doubling in size and then 8 yet again, you're sitting at 32,000 pounds, then another 3-4 and you're sitting anywhere from 96,000 - 128,000 pounds. (I'd lean towards 128,000 since Cloud is standing a good 10-15 feet from the snake). (More exact measurement) So, I booted up GIMP to get an exact measurement for Cloud's height and then copy+pasted the bar I made over and over again over the snake. It is roughly 13 Clouds in length. Cloud is 5'7" or 173 cm in height. 173 x 13 = 2249 cm or 73.79 feet in length. This is bearing in mind Cloud is closer to the camera than the Zolom and so taller in comparison to the thing than he would be if he stood next to it. Likewise the snake is 0.78 Clouds thick, 135.16 cm, or 4.5 feet. This is, again, bearing in mind that Cloud is standing some 10-15 feet away from the thing; hence my earlier 5-6 feet estimate. The rest is just Square Cube Law. Every time something doubles in seize you multiply the weight by 8, no? 2B's speed (Nier Automata) Now, I'll just ballpark this because 2B's FTE finishers and powers still operate as such when Overclock is being used, but hypothetically: The Mirage Pod Program states it locates enemy weaknesses, Analyzes all enemies in a fixed radius, allowing combatants to deliver a series of deadly slashes to all who are caught within. the FTE speeds still come from 2B, she has several finishing moves she can use on enemies that involve her moving FTE without said program. Mirage allows 2B to cut every enemy within I want to say 5 meters of her location at FTE speeds. At least with the max level. As this will hit every enemy within 180 degrees and the description seems to imply she will slash every opponent, I would suggest her running the circumference of the circle within the span of time insisted for FTE movements to be a viable method for getting a speed result, other alternatives exist but I have to start somewhere. With the 5 meters used for radius, the circumference would be 31.42 meters. I'll use 1/60th of a second because it's a 60 FPS game, you'd be free to use other values if you wished. To travel 31.42 meters in 1/60th of a second would be 1885.2 meters per second. Mach 5.54. If you want 1/30th of a second you would cut the result in half. If you want to use 1/220th (I don't) you would multiply the result by something like 3 and a half times for a 250% increase. But what about Overclock? This is where it gets messy as you have to calc-stack. Looking at this video https://www.youtube.com/watch?v=Ajx5NhIiB44 In less than a second, a punch from one of those robots will cover at least 3 meters in normal time. When Overclocked, they will take about one second to cover even 1 meter. (You're free to find some other point of interest for gauging the time dilation in said video). That would mean you could triple 2B's speed to Mach 16.62 if you went with the 60 FPS use of FTE. It's a fairly hypersonic speed but as it's basically multiplying the speed from FTE movements, getting a result above Mach 5 is frankly to be expected. Considering they DO fly from near the moon and back in flight gear and fight thousands of machines in said flight gear while flying around in atmosphere at times, hypersonic speeds and matching reaction times are not exactly unexpected. Edit: It occured to me that she has a pod program that slows time as well. While not applicable for this match, catching an opponent in it would theoretically increase her relative speed even further against the stopped opponent. Assuming only a 50% slow, if it landed, she could reach well over mach 30 using the above values. Mach 15+ under her own power, and her opponent would be twice as slow so she would appear mach 30. Haruko Plays With Bullets (FLCL) The series is fairly nonsensical so consistent feats are a tad difficult to ask for- especially with how hard they lean on the 4th wall. That said, one feat that can give fairly hard numbers is the battle in episode 5 where Haruko takes on a small army of government goons. The battle in question. Seeing as Haruko earlier in the episode swatted a sniper round out of the air- it's hardly unfair to say she's a bullet-timer to begin with. But this battle takes it to a new level. If you actually watch frame-by-frame you can see Haruko sticking her tongue out and dodging the first shot. The problem is that we can't accurately judge where the bullet is and the like to get a speed feat. Later however, where she runs through a swarm of them, we most certainly can. The standard procedure of eyeball first, get specifics later indicates she moves something like 10 feet before the bullets move even 1 foot. As we see all manner of guns being fired from the henchmen, including AK-47s. The firing speed of an AK bullet is 715 meters per second, just barely above Mach 2 at 2.08x the speed of sound. That would result in an easy Mach 20 combat speed when you factor in she deliberately moved through all the bullets AND she wasn't going straight forwards in this timeframe. Category:Blog posts